Question

Use the Laplace transform to solve the given initial value problem.

y^(4)−256y=0; y(0)=43, y′(0)=104, y′′(0)=432, y′′′ (0)=1,024


Enclose arguments of functions in parentheses. For example, sin(2x).

y(t)= _______________

Answer 1

$y(t)\to Y(p)$

$y^{(4)}(t)\to p^4Y(p)-p^3y(0)-p^2y'(0)-py''(0)-y^{(3)}(0)$

For our case:

$y^{(4)}(t)\to p^4Y(p)-43p^3-104p^2-432p-1024$

Then:

$p^4Y(p)-43p^3-104p^2-432p-1024-256Y(p)=0$

$(p^4-256)Y(p)=43p^3+104p^2+432p+1024$

$Y(p)=\frac{43p^3+104p^2+432p+1024}{(p^2+16)(p+4)(p-4)}$

$\frac{A}{p+4}+\frac{B}{p-4}+\frac{Cp+D}{p^2+16}=\frac{43p^3+104p^2+432p+1024}{(p^2+16)(p+4)(p-4)}$

$A(p-4)(p^2+16)+B(p+4)(p^2+16)+(Cp+D)(p+4)(p-4)=$

$=43p^3+104p^2+432p+1024$

$A(p^3-4p^2+16p-64)+B(p^3+4p^2+16p+64)+Cp(p^2-16)+D(p^2-16)=$

$=43p^3+104p^2+432p+1024$

$A+B+C=43$

$-4A+4B+D=104$

$16A+16B-16C=432\implies A+B-C=27$

$-64A+64B-16D=1024\implies -4A+4B-D=64$

$2C=16\implies C=8$

$2D=40\implies D=20$

$A+B=35$

$-4A+4B=84\implies B-A=21$

$B=28,A=7$

Then:

$Y(p)=7\cdot\frac{1}{p+4}+28\cdot\frac{1}{p-4}+\frac{8p+20}{p^2+16}$

$\frac{1}{p+4}\to e^{-4t}$

$\frac{1}{p-4}\to e^{4t}$

$\frac{p}{p^2+16}\to cos4t$

$\frac{1}{p^2+16}\to \frac{1}{4}sin4t$

Answer:

$y(t)=7e^{-4t}+28e^{4t}+8cos4t+5sin4t$